I want to know whether the image of a convex polytope $\mathcal{M} \subseteq R^{n \times n}$ under the mapping $A \mapsto exp(\tau A)$ remains a convex polytope (not necessarily $\mathcal{M}$)? Here $A \in \mathcal{M}(R^{n \times n})$ and $\tau > 0$ is known constant.
Further, what if the map is $A \mapsto Aexp(\tau A)$?
Thanks in advance. Structured explanation would help.
No, consider the line segment between $\pmatrix{1&0\\0&0}$ and $\pmatrix{0&0\\0&1}$ and let $\tau=1$. The image includes $\pmatrix{e&0\\0&1}$ and $\pmatrix{1&0\\0&e}$, but not the line segment between them, such as $\pmatrix{\frac{e+1}{2}&0\\0&\frac{e+1}{2}}$.
Edit: More explicitly: Let $\mathcal{M}$ be the line segment between $a=\pmatrix{1&0\\0&0}$ and $b=\pmatrix{0&0\\0&1}$. Let $f:A\mapsto\exp(A)$. Then $f(\mathcal{M})$ is not convex. Because: $$f(a)=\pmatrix{e&0\\0&1}\in f(\mathcal{M})$$ $$f(b)=\pmatrix{1&0\\0&e}\in f(\mathcal{M})$$ But $$\frac{f(a)+f(b)}{2}=\pmatrix{\frac{e+1}{2}&0\\0&\frac{e+1}{2}}\not\in f(\mathcal{M}).$$